Evaluate $\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin x} \text{d}x$

Question

Evaluate $$ \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin x} \text{d}x$$
where $n\in\mathbb{N}$

I was able to do this using parametization (differentiation under integration) but I'm not allowed to use that in my school since it isn't in the syllabus :(
I should only use high school level mathematics. However, I can't see a way to do this with an elementary approach.

Answer 1

Recall the identity: $$\sin^2(A) - \sin^2(B) = \sin(A+B)\sin(A-B)$$ Let $I_n = \displaystyle \int_0^{\pi/2} \dfrac{\sin^2(nx)}{\sin(x)}dx$, we obtain \begin{align} I_{n+1} - I_n & = \int_0^{\pi/2} \dfrac{\sin^2((n+1)x)}{\sin(x)}dx - \int_0^{\pi/2} \dfrac{\sin^2(nx)}{\sin(x)}dx\\ & = \int_0^{\pi/2}\dfrac{\sin^2((n+1)x)-\sin^2(nx)}{\sin(x)} dx\\ & = \int_0^{\pi/2}\dfrac{\sin((2n+1)x)\sin(x)}{\sin(x)} dx\\ & = \int_0^{\pi/2} \sin((2n+1)x)dx\\ & = -\left.\dfrac{\cos((2n+1)x)}{2n+1} \right \vert_0^{\pi/2}\\ & = \dfrac1{2n+1} \end{align} We have $I_0 = 0$ and hence $$I_{n} = \dfrac11 + \dfrac13 + \dfrac15 + \cdots + \dfrac1{2n-1}$$

Answer 2

Hint:

Start by using the power reduction formula to get rid of the sine-squared term:

$$\begin{align} \mathcal{I}_{n} &=\int_{0}^{\frac{\pi}{2}}\frac{\sin^2{nx}}{\sin{x}}\,\mathrm{d}x\\ &=\int_{0}^{\frac{\pi}{2}}\frac{1-\cos{2nx}}{2\sin{x}}\,\mathrm{d}x.\\ \end{align}$$

Then, writing the $\cos{2nx}$ as a binomial series, we can rewrite the integrand as a polynomial of sines and cosines:

$$\cos{(2nx)}=\sum_{k=0}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$

$$\implies 1-\cos{(2nx)}=1-\sum_{k=0}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$

$$\implies 1-\cos{(2nx)}=1-\cos^2{(x)}-\sum_{k=1}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$

$$\implies 1-\cos{(2nx)}=\sin^2{(x)}-\sum_{k=1}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$

$$\implies \frac{1-\cos{(2nx)}}{\sin{(x)}}=\sin{(x)}-\sum_{k=1}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k-1}{(x)}.$$

Finally, integrate term-by-term.